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So I was asked to prove the following term is equal to $2016$:

$$ \left( \frac{251}{ \frac{1}{ \sqrt [3] {252} - 5 \sqrt [3] {2} } -10 \sqrt [3] {63} } + \frac {1} { \frac {251} { \sqrt [3] {252} +5 \sqrt [3] {2} } + 10 \sqrt [3] {63} } \right)^3 $$

For the record, I know how to solve the question; defining $a^3=252 $ and $b^3 = 250 $ simplifies the expression so it can be solved.

But I was hoping (and this may be too much to hope for) that there was some reason why this radical expression simplifies so nicely, and would give a straightforward way of deriving the problem. So is there anything special to this expression?

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  • $\begingroup$ May have to do with nested radicals, but I'm not sure. $\endgroup$ – Simply Beautiful Art Jul 20 '16 at 1:10
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This is related to your approach, but possible easier. Hard to tell since you didn't give your argument.

$$\begin{align} f(x,y)&=\frac{\frac{1}{2}(x^3+y^3)}{\frac{\frac{1}{2}(x^3-y^3)}{x-y}-xy}\\ &=\frac{x^3+y^3}{\frac{x^3-y^3}{x-y}-2xy}\\ &=\frac{x^3+y^3}{x^2+xy+y^2-2xy}\\ &=\frac{x^3+y^3}{x^2-xy+y^2}\\ &=x+y \end{align}$$

Then $f(x,y)+f(x,-y)=2x$.

In the case of $x=\sqrt[3]{252},y=\sqrt[3]{250}=5\sqrt[3]{2}$, $\frac{x^3+y^3}{2}=251, \frac{x^3-y^3}{2}=1, xy=10\sqrt[3]{63}$.

This also lets you see that:

$$\left( \frac{251}{ \frac{1}{ \sqrt [3] {252} - 5 \sqrt [3] {2} } -10 \sqrt [3] {63} } - \frac {1} { \frac {251} { \sqrt [3] {252} +5 \sqrt [3] {2} } + 10 \sqrt [3] {63} } \right)^3=2000$$

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  • $\begingroup$ My question was really more along the lines of whether the fact this expression simplifies to an integer is due to some more fundamental result in mathematics. Perhaps I should also clarify that, by derive, I mean how this question was created. You could create it by going backwards with the steps you've shown in your answer, but I was hoping that this question was just a special case in a more general concept (which is a lot to hope for). $\endgroup$ – Cataline Jul 20 '16 at 0:06
  • $\begingroup$ It was almost certainly created by taking $x+y=\frac{x^3+y^3}{x^2-xy+y^2}$ and proceeding from there, using that $x^2-xy+y^2=(x^2+xy+y^2)-2xy$. Basically, there isn't anything deep here. $\endgroup$ – Thomas Andrews Jul 20 '16 at 0:09

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